Answer:
what is it?? plss helpp taking exam rn
Explanation:
Answer:
21.2 gm
Explanation:
calculate the mass of butane needed to produce 64.1 g of carbon dioxide to three significant figures and appropriate units
butane is the hydrocarbon C4H10
in combustion, we react hydrocarbons with O2 to form CO2 and H2O
so
C4H10 + O2----------------> CO2 + H2O
BALANCE
2C4H10 + 1302--------> 8CO2 + 10 H2O
the molar mass of CO2 is 12 + 16X2 = 44
64.1 gm of CO2 is
64.1/44 = 1.46 MOLES OF CO2,
FOR EVERY 8 MOLES OF CO2 WE NEED 2 MOLES OF BUTANE IT IS A
8:2 OR 4:1 RATIO. THE MOLES OF C4H10 ARE 1/4 THE MOLES OF CO2
SO
THE MOLES OF C4H10 H10 ARE 1.46/4 =0.365 MOLES
THE MOLAR MASS OF BUTANE IS 58.12
0.365 MOLES OF C4H10 HAS A MASS OF 0.365 X 58.12 = 21.2 gm
Answer:
The answer to your question is A. Gases are not easily compressed because of expansion and limited space between particles.
Answer:
2.5x10^–3 mole.
Explanation:
Data obtained from the question include:
Volume of solution = 25mL
Molarity of HNO3 = 0.1M
Mole of HNO3 =..?
First, we'll begin by converting 25mL to L. This can be achieved by doing the following:
1000mL = 1L
Therefore, 25mL = 25/1000 = 0.025L
Now, we can obtain the number of mole of HNO3 present in the solution as follow:
Molarity = mole /Volume
Mole = Molarity x Volume
Mole = 0.1 x 0.025
Mole = 2.5x10^–3 mole.
Therefore, 2.5x10^–3 mole of HNO3 is present in the solution.
